2 Answers
2

The condition $\lambda>1$ is sufficient and, at least almost, necessary:

To clarify, the space of irrational numbers $(0,1)-\mathbb Q$ is homeomorphic to
$\omega^\omega$ under the map that sends $\frac{1}{a_1+\frac{1}{a_2+\cdots}}$ to a function $f$ satisfying
$f(n)=a_{n+1}-1$. (As is well known.)

This way Lebesgue measure on $(0,1)$ induces a "Gaussian" measure on $\omega^\omega$.

Under this measure $\mu$, I claim that the following set has measure one:
$$\{f:f(n)\text{ is eventually bounded by $n^t$ for $t>1$, but not for
$t=1$}\}$$

Proof:
With $\delta_n=\sum_{k\le n^t}\epsilon_{nk}$ and
$\Delta_n=\delta_n\log 2$ and $\log=\log_e$, we have

Thank you very much. Could you suggest me books regarding these facts.
–
sokhoJul 16 '14 at 23:07

Actually my coauthor Bonnie S. Huggins and I just proved it from scratch. But there should be some books about continued fractions and the Gaussian measure on the irrational numbers.
–
Bjørn Kjos-HanssenJul 16 '14 at 23:12

A precise solution to this problem is known. In Khintchine's book on continued fractions it says:

$\mathbf{Theorem~30}$ Suppose that $\varphi(n)$ is an arbitrary positive function with natural argument $n$. The inequality
$$
a_n = a_n(\alpha) \geq \varphi(n)
$$
is, for almost all $\alpha$, satisfied by an infinite number of values $n$ if the series $\sum_n 1/\varphi(n)$ diverges. On the other hand, this inequality is, for almost all $\alpha$, satisfied by only a finite number of values of $n$ if the series $\sum_n 1 /\varphi(n)$ converges.